Nuclear Physics: Question 6

Worked solution

Step 1: Recall the relationship between activity, decay constant and number of nuclei

Activity is the rate at which nuclei decay, given by: $A=\lambda N$

Step 2: Substitute the values

$A=5.08\times10^{-11}\times3.00\times10^{14}$

Multiply the mantissas and add the powers of ten separately: $5.08\times3.00=15.24,\qquad10^{-11}\times10^{14}=10^{3}$

$A=15.24\times10^{3}=1.524\times10^{4}\text{ Bq}$

Recompute as a check, grouping the powers of ten differently: $A=(5.08\times10^{3})\times(3.00\times10^{0})\times10^{0}$… more simply, swapping which factor carries which exponent, $A=(5.08\times10^{-11}\times10^{14})\times3.00=(5.08\times10^{3})\times3.00=15.24\times10^{3}\text{ Bq}$, the same result.

$A=\boxed{1.52\times10^{4}}\text{ Bq (3 s.f.)}$

Why the other options are wrong

• B ($1.52\times10^{-10}\text{ Bq}$): comes from forgetting to include the $10^{14}$ from $N$, leaving only $5.08\times3.00\times10^{-11}$.
• C ($5.91\times10^{24}\text{ Bq}$): comes from rearranging the formula incorrectly and computing $N/\lambda$ instead of $\lambda N$.
• D ($1.52\times10^{13}\text{ Bq}$): comes from an arithmetic slip when combining the powers of ten, e.g. treating $10^{-11}\times10^{14}$ as $10^{13}$ instead of $10^{3}$.

Final answer

• The activity of the source is $\boxed{1.52\times10^{4}}\text{ Bq}$, option A.